How can I solve the Linear Harmonic Oscillator in polar coordinates?

[Nemanja989]

Hello there,

Can anyone help me, I am struggling with solving LHO in two dimension,but in the polar coordinates.
I transfer laplacian into polar from decart coordinates, write Ψ=ΦR, and do Fourier separation method for solving differential equation. But I do not know how to solve differential equation on R.

I will write to where I came tomorrow, or in a few hours.

Thanks!

 

[Jivesh]

Why trouble yourself by going into polar co-ordinates? Just look for solution of the form Ψ = X(x)Y(y), for the Schrodinger equation in cartesian co-ordinates and you'll get two equations for a Linear Harmonic Oscillator. Surely, for the case of two one dimensional equation, you know the solution, multiply them, and add there energy levels to get the answer. I think the lowest energy state would have the energy of (hbar)*w, where w is the natural frequency of oscillation and the wavefunction is a Gaussian in radial co-ordinate r.

 

[Dickfore]

In this problem, both \phi and z are cyclic coordinates (they do not enter explicitly in the Schroedinger equation), so the corresponding conjugate momenta p_{\phi} \equiv l_{z} and p_{z} commute with the Hamiltonian. Therefore, the stationary states of the system are of the form:

<br /> \psi_{m, p_{z}}(\rho, \phi, z) = \frac{1}{(2 \pi \hbar)^{1/2}} \, e^{\frac{i}{\hbar} \, p_{z} \, z} \, \frac{1}{\sqrt{2 \, \pi}} \, e^{i \, m \, \phi} \, R_{m}(\rho), \ m = \ldots, -1, 0, 1, \ldots<br />

Substitute this into the Schroedinger equation and you will get the following oridnary differential equation:

<br /> -\frac{\hbar^{2}}{2 \, \mu} \, \left[ \frac{1}{\rho} \, \frac{d}{d \rho}\left(\rho \, \frac{dR_{m}(\rho)}{d \rho}\right) - \frac{m^{2}}{\rho^{2}} \, R_{m}(\rho) - \frac{p^{2}_{z}}{\hbar^{2}} \, R_{m}(\rho) \right] + \frac{\mu \, \omega^{2} \, \rho^{2}}{2} \, R_{m}(\rho) = E \, R_{m}(\rho)<br />

The normalization condition is:

<br /> \int_{0}^{\infy}{\rho \, R^{2}_{m}(\rho) \, d\rho} = 1<br />

Introduce a dimensionless argument:

<br /> \rho = a x, R_{m}(\rho) = \frac{1}{a} \, y_{m}(x)<br />

with:

<br /> a = \left(\frac{\hbar}{\mu \, \omega}\right)^{\frac{1}{2}}<br />

and:

<br /> \epsilon = \frac{2}{\hbar \, \omega} \, \left( E - \frac{p^{2}_{z}}{2 \, \mu} \right)<br />

you will get the following equation:

<br /> y&#039;&#039;_{m} + \frac{1}{x} \, y&#039;_{m}(x) + \left(\epsilon - x^{2} - \frac{m^{2}}{x^{2}}\right) \, y_{m}(x) = 0<br />

with the normalization condition:

<br /> \int_{0}^{\infty}{x \, y^{2}_{m}(x) \, dx} = 1<br />

Usually, we want to get rid of the first derivative in the differential equation. We can achieve this by the substitution:

<br /> y_{m}(x) = \frac{z_{m}(x)}{\sqrt{x}}<br />

A direct substitution should convince you that the equation satisfied by z_{m}(x) is:

<br /> z&#039;&#039;_{m} + \left(\epsilon - x^{2} - \frac{m^{2} - 1/4}{x^{2}}\right) \, z_{m} = 0<br />

and the normalization condition reads:

<br /> \int_{0}^{\infty}{z^{2}_{m}(x) \, dx} = 1<br />

The asymptotic behavior for z_{m}(x) for both large and small values of x is obtained by keeping the dominant terms in the equation for the relevant region:

<br /> z&#039;&#039;_{m 0} - \frac{m^{2} - 1/4}{x^{2}} \, z_{m 0} = 0, \ x \rightarrow 0<br />

This is an Euler equation and has solutions of the form z_{m 0} \tilde x^{k}. Substituting, we get an algebraic equation for k:

<br /> k^{2} - k - \left(m^{2} - \frac{1}{4}\right) = 0<br />

the solutions of which are:

<br /> k_{1/2} = \frac{1 \pm 2 \, |m|}{2}<br />

We take:

<br /> z_{m 0} \sim x^{|m| + \frac{1}{2}}, \ x \rightarrow 0<br />

For large x, we have:

<br /> z&#039;&#039;_{m \infty} - x^{2} \, z_{m \infty} = 0<br />

which is of the same form as in the one dimensional case, so we have:

<br /> z_{m \infty} \sim e^{-\frac{x^{2}}{2}}, \ x \rightarrow \infty<br />

Finally, we capture the asymptotic behavior by writing:

<br /> z_{m}(x) = x^{|m| + 1/2} \, e^{-x^{2}/2} \, v_{m}(x)<br />

After some differentiation and algebraic manipulation, you should get the following equation for v_{m}(x):

<br /> v&#039;&#039;_{m} + \left(\frac{2 |m| + 1}{x} - 2 x \right) \, v&#039;_{m} + \lambda \, v_{m} = 0<br />

where

<br /> \lambda = \epsilon - 2 |m| - 1<br />

Using a change of variables in the argument:

<br /> x = a \, t^{\alpha}<br />

you will get:

<br /> t \, \ddot{v}_{m} + \left[ 1 + 2 \, \alpha \, |m| - 2 a^{2} \, \alpha \, t^{2 \, \alpha} \right] \, \dot{v}_{m} + \lambda \, (a \alpha)^{2} t^{2 \, \alpha - 1} = 0

Compare this with the differential equation for the Kummer confluent hypergeometric equation _{1}F_{1}(a; c; t):

<br /> t \ddot{y} + (c - t) \, \dot{y} - a \, y = 0<br />

we see that we should have:

<br /> 2 \, \alpha = 1, \; 2 \, \alpha \, a^{2} = 1 \Rightarrow \alpha = \frac{1}{2}, \; a = 1 \Rightarrow x = t^{\frac{1}{2}} \Leftrightarrow t = x^{2}<br />

So, we can write:

<br /> v_{m}(x) = _{1}F_{1}(-\frac{\lambda}{4}, |m| + 1, x^{2})<br />

Collecting everything back, the radial wavefunction is:

<br /> R_{n m}(a \, x) = C_{n, m} \, x^{|m|} \, e^{-\frac{x^{2}}{2}} \, _{1}F_{1}\left( -n, |m| + 1, x^{2} \right), n \in \mathbb{N}_{0}, m \in \mathbb{Z}<br />

where C_{n, m} is a normalization constant. The energy eigenvalues associated with motion in the plane of the oscillator are:

<br /> E_{n, m, p_{z}} - \frac{p^{2}_{z}}{2 \, \mu} = E&#039;_{n, m} = \hbar \, \omega \, \left( |m| + 2 \, n + \frac{1}{2} \right)<br />

 

[Nemanja989]

I have attached a pdf file into this post.

Thank you for responding!

You asked me why I do it in polar coordinates, when it`s much easier in decarte. Well I simply want to practice on my own, I want to solve one problem in a few ways. I have just finished basic cousre of quantum mechanics, but we haven`t done LHO in polar coordinates.